From Uniform Convergence to a Function Space
Uniform convergence compares functions by controlling their differences at every point of a domain. To use that viewpoint systematically, it helps to identify a collection of functions that stays within the same class under familiar operations. Bounded real-valued functions provide a natural starting point: their values remain within some finite range, and pointwise sums, scalar multiples, and products remain bounded as well.
The collection depends on the domain. A function can be bounded on one set and unbounded on another, and boundedness alone does not imply continuity. We will define the space for a fixed nonempty set \(E\), establish its algebraic structure, and check these distinctions with examples. The notation \(B(E)\) has already appeared in this course; here we focus on the meaning and structure of that space.
The Space of Bounded Functions
A real-valued function is bounded when its values have a common finite bound in absolute value. The bound is allowed to depend on the function and on its domain; it need not be attained by the function.
The condition asks for one constant that works for every point in \(E\). It does not ask for a different bound at each point. When \(E\) is fixed, we can compare functions in \(B(E)\) by using the usual pointwise operations:
Here \(f,g:E\to\mathbb{R}\), \(c\in\mathbb{R}\), and \(x\in E\). These formulas define new functions on the same domain. We next check that, when the original functions are bounded, the resulting functions remain in \(B(E)\).
Proof. Since \(f\) and \(g\) are bounded, there are finite \(A,B\geq0\) such that \(|f(x)|\leq A\) and \(|g(x)|\leq B\) for every \(x\in E\). The triangle inequality gives, for every \(x\in E\), $$ |(f+g)(x)|\leq |f(x)|+|g(x)|\leq A+B. $$ Thus \(f+g\) is bounded. Also, $$ |(cf)(x)|=|c|\,|f(x)|\leq |c|A $$ for every \(x\in E\), so \(cf\) is bounded, including when \(c=0\). For the product, $$ |(fg)(x)|=|f(x)|\,|g(x)|\leq AB $$ for every \(x\in E\), so \(fg\) is bounded. Finally, the constant function \(1\) satisfies \(|1|\leq1\) at every point and is therefore bounded. \(\square\)
This theorem says more than that certain formulas make sense pointwise: it says that those operations do not take us outside \(B(E)\). The zero function also belongs to \(B(E)\), since its absolute value is bounded by \(0\). Therefore \(B(E)\) is a real vector space under pointwise addition and scalar multiplication. The vector-space identities hold because they hold for real numbers at each \(x\in E\). For example, \(f+g=g+f\) follows from \(f(x)+g(x)=g(x)+f(x)\) for every \(x\).
Pointwise multiplication adds a compatible algebraic operation. It is associative and commutative, has the constant function \(1\) as a multiplicative identity, and distributes over addition. Each identity follows by evaluating both sides at any \(x\in E\) and using the corresponding real-number identity. In this sense, \(B(E)\) is a commutative algebra over \(\mathbb{R}\): it is a real vector space with an associative, commutative multiplication that is compatible with the vector-space operations.
Worked Example: An Explicit Bound for a Sum and Product
Let \(E=[-2,2]\), \(f(x)=x^2-1\), and \(g(x)=3-x\). Since \(0\leq x^2\leq4\), we have \(-1\leq f(x)\leq3\), and hence \(|f(x)|\leq3\). Also, \(1\leq g(x)\leq5\), so \(|g(x)|\leq5\). Therefore, the theorem gives the bounds \(|f(x)+g(x)|\leq8\) and \(|f(x)g(x)|\leq15\) on all of \(E\).
The formulas can be checked directly: $$ (f+g)(x)=x^2-x+2,\qquad (fg)(x)=(x^2-1)(3-x). $$ The bounds above apply to these functions because they are sums and products of the bounded functions \(f\) and \(g\). They need not be optimal; their purpose is to supply a single finite bound that works throughout the domain.
The Domain Changes the Boundedness Question
Boundedness is a property of a function together with its domain. The same formula can define a bounded function on one domain and an unbounded function on another. One useful consequence is that finite domains are especially simple: there are only finitely many function values to control.
Proof. Write \(E=\{x_1,\ldots,x_m\}\), where \(m\geq1\). The finite list \(|f(x_1)|,\ldots,|f(x_m)|\) has a largest value; call it \(M\). For each \(x\in E\), there is an index \(j\) with \(x=x_j\), so $$ |f(x)|=|f(x_j)|\leq M. $$ Thus one finite bound works at every point of \(E\), and \(f\in B(E)\). \(\square\)
Worked Example: Every Function on a Three-Point Set Is Bounded
Let \(E=\{a,b,c\}\), and suppose \(f(a)=-7\), \(f(b)=2\), and \(f(c)=5\). The largest absolute value among these three function values is \(7\), so $$ |f(a)|=7\leq7,\qquad |f(b)|=2\leq7,\qquad |f(c)|=5\leq7. $$ Thus \(f\) is bounded on \(E\). No continuity or other regularity property is needed. The finite-domain theorem applies to any real-valued assignment to \(a,b,c\), not just these particular values.
On an infinite domain, the definition must be checked rather than inferred from the fact that each individual value is a real number. Having a finite value at every point does not ensure that the values share a common finite bound.
Worked Example: A Bounded Function on the Real Line
Define \(h:\mathbb{R}\to\mathbb{R}\) by $$ h(x)=\frac{x}{1+x^2}. $$ For every real \(x\), the square \((|x|-1)^2\) is nonnegative. Expanding it gives $$ x^2-2|x|+1\geq0, $$ so \(2|x|\leq1+x^2\). Because \(1+x^2>0\), division by \(2(1+x^2)\) yields $$ |h(x)|=\frac{|x|}{1+x^2}\leq\frac12. $$ The same bound works for every \(x\in\mathbb{R}\), proving that \(h\in B(\mathbb{R})\). This argument establishes boundedness directly, without requiring the function to attain its largest absolute value.
Boundedness Does Not Mean Continuity
The space \(B(E)\) is defined by a bound on values, not by a condition on how those values change as the input changes. It therefore includes functions with jumps and other discontinuities. Conversely, even a continuous function on an unbounded domain need not be bounded. Keeping these properties separate prevents a common error: applying a conclusion that needs boundedness merely because a function is continuous, or assuming that boundedness supplies continuity.
Worked Example: A Bounded Function with a Discontinuity
Define \(q:\mathbb{R}\to\mathbb{R}\) by $$ q(x)= \begin{cases} 1,&x\geq0,\\ -1,&x<0. \end{cases} $$ For every real \(x\), \(|q(x)|=1\), so \(q\in B(\mathbb{R})\). However, \(q\) is not continuous at \(0\): for every \(x<0\), \(q(x)=-1\), while \(q(0)=1\). In particular, taking negative inputs arbitrarily close to \(0\) gives function values that remain \(2\) away from \(q(0)\). Thus boundedness does not imply continuity.
The reverse distinction is visible on unbounded domains. The function \(r:\mathbb{R}\to\mathbb{R}\), \(r(x)=x\), is continuous, but it is not bounded: for any proposed finite bound \(M\geq0\), choosing \(x=M+1\) gives \(|r(x)|=M+1>M\). The finite-domain theorem does not apply because \(\mathbb{R}\) is not finite. Continuity and boundedness answer different questions, and neither should be substituted for the other without an appropriate theorem.
What the Space Structure Gives Us
Once bounded functions are collected in \(B(E)\), a proof can use closure under pointwise operations without repeatedly treating each output as a new boundedness problem. If \(f,g\in B(E)\), then any expression built from finitely many additions, scalar multiplications, and products of \(f\) and \(g\) again defines a bounded function. This follows by applying the Pointwise Operations on Bounded Functions Theorem at each operation.
For example, every polynomial expression \(a_0+a_1f+\cdots+a_m f^m\), with real coefficients \(a_0,\ldots,a_m\), is bounded when \(f\in B(E)\). Indeed, the constant functions are bounded; repeated applications of the product result show that each power \(f^k\) is bounded; and the sum and scalar-multiple results then give boundedness of the entire expression. This conclusion concerns finite algebraic expressions. It does not, by itself, give a bound for an infinite sum or a limit; those require additional hypotheses and results.
The space \(B(E)\) is also the setting for the supremum norm and for the completeness result established elsewhere in this course. Those results use the same underlying collection of functions, but they add structure beyond the algebraic closure proved here. For now, the essential point is that the fixed-domain class \(B(E)\) is stable under the basic pointwise operations and contains all functions on a finite domain, while allowing discontinuous functions on infinite domains.
Check Your Understanding
Use the definition and closure results to answer the following questions.
- If \(|f(x)|\leq A\) and \(|g(x)|\leq B\) on \(E\), what bounds follow for \(f+g\) and \(fg\)?
- Why does every real-valued function on a nonempty finite set belong to \(B(E)\)?
- Can a bounded function be discontinuous? Give the example from this tutorial and identify its bound.
- Why does continuity of \(r(x)=x\) on \(\mathbb{R}\) not imply that \(r\) belongs to \(B(\mathbb{R})\)?
- Why does closure under finite sums and products not, on its own, establish that an infinite series of bounded functions has a bounded sum?